Answer:
0.0446 moles of NaCl are produced by the reaction
Explanation:
First of all we convert the mass of reactants to moles, to determine the limiting reagent
3 g / 134.45 g/mol =0.0223 moles of CuCl₂
4 g / 85 g/mol = 0.0470 moles of NaNO₃
The reaction is: CuCl₂(aq) + 2NaNO₃(aq) → 2NaCl(aq) + Cu(NO₃)₂(aq)
By stoichiomety, 1 mol of chloride reacts with 2 moles of nitrate
Then, 0.0223 moles of chloride will react with (0.0223 . 2) / 1 = 0.0446 moles of nitrate. We have 0.0470 moles, so the nitrate is in excess, and the limiting reactant is the chloride.
2 moles of nitrate react with 1 mol of chloride
Then, 0.0470 moles of nitrate will react with (0.0470 . 1) / 2 = 0.0235 moles of chloride. There is not enough chloride because we have 0.0223 moles and we need 0.0235.
In the reaction, 1 mol of chloride produces 2 mol of sodium chloride
Then, our 0.0223 moles will produce (0.0223 .2 ) / 1 = 0.0446 moles of NaCl
